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2 years ago

How do iquantum tunel through a wall and how long will it take me?

Physics

1 answer:

Anon25 [30]2 years ago

5 0

Answer:

The phenomenon known as "tunneling" is one of the best-known predictions of quantum physics, because it so dramatically confounds our classical intuition for how objects ought to behave. If you create a narrow region of space that a particle would have to have a relatively high energy to enter, classical reasoning tells us that low-energy particles heading toward that region should reflect off the boundary with 100% probability. Instead, there is a tiny chance of finding those particles on the far side of the region, with no loss of energy. It's as if they simply evaded the "barrier" region by making a "tunnel" through it.

Explanation:

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A 60 kg skier starts from rest at the top of a frictionless slope of height of 35 meters. The velocity at the bottom is v. If a

Solnce55 [7]

Answer:

Explanation:

Speed of skier without parachute

= √ 2gh

= √ 2 x 9.8 x 35

= 26.2 m / s

Speed of skier with parachute

net force downwards

mg - 200

= 60 x 9.8 -200

= 388 N

acceleration = 388 / 60

a = 6.47 m / s

v = √ 2ah

= √ 2 x 6.47 x 35

= 21.28 m / s

8 0

2 years ago

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Physics Help Please!!!

liq [111]

The answer is the third graph

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2 years ago

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The force involved in falling of an apple from a tree is known as? (a) Magnetic force (b) electrostatic force (c) Contact force

musickatia [10]

Answer:

gravitational force

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6 0

2 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$